Question

youre evaluating the number of books borrowed by students from the school library over a month to determine reading habits. the number of books borrowed per student is as follows:
\\{1, 2, 2, 2, 3, 4, 5, 5, 7, 12\\}.

should you use a normal curve to represent this data? why or why not?

• yes, the data is symmetrically distributed, and the mean, median, and mode are close, indicating a normal curve is appropriate.
• yes, the data has no significant outliers, and the values are spread around the mean, making it suitable for a normal curve.
• no, the data set is too small to accurately use a normal curve.
• no, the data is right-skewed, with a few high values (outliers) pulling the mean upward, so it does not fit a normal curve.

Explanation:

Calculate central tendency measures

We analyze the dataset:

$$ \{1, 2, 2, 2, 3, 4, 5, 5, 7, 12\} $$

The number of data points is \(n = 10\).
The mean is:

$$ \mu = \frac{1+2+2+2+3+4+5+5+7+12}{10} = \frac{43}{10} = 4.3 $$

The median (average of the 5th and 6th values, 3 and 4) is:

$$ \text{Median} = \frac{3 + 4}{2} = 3.5 $$

The mode (most frequent value) is \(2\).

Analyze distribution shape

We compare the calculated measures:

$$ \text{Mode} (2) < \text{Median} (3.5) < \text{Mean} (4.3) $$

This ordering indicates a right-skewed distribution.
The value \(12\) is much larger than the rest, acting as an outlier that pulls the mean upward.

Evaluate normal curve suitability

A normal distribution must be symmetric.
Since this dataset is strongly right-skewed, a normal curve is not appropriate.
This matches the fourth option.

Answer:

• (A) Yes, the data is symmetrically distributed, and the mean, median, and mode are close, indicating a normal curve is appropriate.
• (B) Yes, the data has no significant outliers, and the values are spread around the mean, making it suitable for a normal curve.
• (C) No, the data set is too small to accurately use a normal curve.
• (D) No, the data is right-skewed, with a few high values (outliers) pulling the mean upward, so it does not fit a normal curve. (Correct answer)